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On [alpha]-[tau]-disconnectedness and [alpha]-[tau]-connectedness in topological spaces/Sobre a [alpha]-[tau]-des- conectividade e a [alpha]-[tau]-conectividade em espacos topologicos.

Introduction

There are several natural approaches that can take to rigorously the concepts of connectedness for a topological spaces. Two most common approaches are connected and path connected and these concepts are applicable Intermediate Mean Value Theorem and use to help distinguish topological spaces. These concepts play a significant role in application in geographic information system studied by Egenhofer and Franzosa (1991), topological modelling studied by Clementini et al. (1994) and motion planning in robotics studied by Farber et al. (2003). The generalization of open and closed set as like [alpha]-open and [alpha]-closed sets was introduced by Njastad (1965) which is nearly to open and closed set respectively. These notion are plays significant role in general topology. In this paper, the new approaches of separate sets, disconnected sets and connected sets called [alpha]-[tau]-separate set, [alpha]-[tau]-disconnected sets and [alpha]-[tau]-connected set with the help of [alpha]-open and [alpha]-closed set are firstly introduced. Further, some relationship concerning [alpha]-[tau]-disconnected and [alpha]-[tau]-connected sets with [alpha]-[tau]-separate sets are also investigated.

Throughout this paper (X,[tau]) and (X,[[tau].sub.[alpha]]) will always be topological spaces. For a subset A of topological space X, Int(A), Cl(A), [Cl.sub.[alpha]](A) and [Int.sub.[alpha]](A) denote the interior, closure, [alpha]-closure and [alpha]-interior of A respectively and [G.sub.[alpha]] is the [alpha]-open set for topology [[tau].sub.[alpha]] on X.

Preliminaries

We shall requires the following definitions and results.

Definition 2.1. Levine (1963), defined a subset A of (X,[tau]) is semi-open if A [subset] Cl(Int(A)) and its complement is called semi-closed set.

The family SO(X,[tau]) of semi-open sets is not a topology on X.

Definition 2.2. Mashhour et al. (1982), defined a subset A of (X,[tau]) is called pre-open locally dense or nearly open if A [subset] Int (Cl (A)) and its complement is called pre-closed set.

Theorem 2.3. According to Mashhour et al. (1982), the family PO(X,[tau]) of pre-open sets is not a topology on X.

Definition 2.4. Maheshwari and Jain (1982), defined a subset A of (X,[tau]) is called [alpha]-open if A [subset] Int(Cl(Int(A))). The family [[tau].sub.[alpha]] of [alpha]-open sets of (X,[tau]) is a topology on X which is finer than T and the complement of an [alpha]-open set is called an [alpha]-closed set.

Theorem 2.5. According to Njastad (1965), for a subset A of (X,[tau]), thee following are equivalent:

A [subset] [[tau].sub.[alpha]];

A is semi open and pre-open;

A [intersection] B [member of] SO(X,[tau]), for all B in So(X,[tau]);

A = O / N, where O [member of] [[tau].sub.[alpha]] and N is nowhere dense;

There exists U contained in T a such that

U [subset] A [subset] scl(U) = Int(Cl(A)).

Theorem 2.6. The intersection of semi-open (resp. pre-open) set and an [alpha]-open set is semi-open (resp. pre-open).

Theorem 2.7. According to Njastad (1965), SO(X,[tau]) = SO(X,[[tau].sub.[alpha]]) and PO(X,[tau]) = PO(X,[[tau].sub.[alpha]]).

Theorem 2.8. According to Njastad (1965), the [alpha]-open sets A and B are disjoint if and only if Int(Cl(Int(A))) and Int(Cl(Int(B))) are disjoint.

Definition 2.9. A point x in X is called an [alpha]-interior point of a set A in X if there exists A [contains] [G.sub.[alpha]] [member of] [[tau].sub.[alpha]] such that y [member of] [G.sub.[alpha]] with y [not equal to] X, i.e. x in X is called an [alpha]-interior point of a set A in X if foe every [G.sub.[alpha]] [member of] [[tau].sub.[alpha]] with x [member of] [G.sub.[alpha]] and [G.sub.[alpha]] [subset] A. Collection of all [alpha]-interior points of A is called [alpha]-interior of A which is denoted by [Int.sub.[alpha]](A). Alternatively, we can define as [Int.sub.[alpha]](A) by [Int.sub.[alpha]](A) = [union]{[G.sup.[alpha]] [member of] [[tau].sub.[alpha]]: [G.sub.[alpha]] [subset] A}.

Main Results

Definition 3.1. Let (X,[tau]) and (X,[[tau].sub.[alpha]]) be topological spaces. Then the subsets A and B of (X,[tau]) are said to be [alpha]-[tau]-separate sets if and only if

A and B are non-empty set.

A [intersection] [Cl.sub.[alpha]](B) and B [intersection] [Cl.sub.[alpha]](A) are non-empty.

Remark 3.2 If A and B are [alpha]-[tau]-separate sets, then both of them are also disjoint sets.

Definition 3.3. Let (X,[tau]) and (X,[[tau].sub.[alpha]]) be topological spaces. Then the subsets A of X is said [alpha]-[tau]-disconnected, if there exists [G.sub.[alpha]] and [H.sub.[alpha]] in [T.sub.[alpha]] such that

A [intersection] [G.sub.[alpha]] and A [intersection] [H.sub.[alpha]] [not equal to] [phi].

(A [intersection] [G.sub.[alpha]]) [intersection] (A [intersection] [H.sub.[alpha]]) = [phi].

(A [intersection] [G.sub.[alpha]]) [union] (A [intersection] [H.sub.[alpha]]) = A.

(X,[[tau].sub.[alpha]]) is said to be [alpha]-[tau]-disconnected if there exists non-empty [G.sub.[alpha]] and [H.sub.[alpha]] in [[tau].sub.[alpha]] such that [G.sub.[alpha]] [intersection] [H.sub.[alpha]] and [G.sub.[alpha]] [union] [H.sub.[alpha]] [not equal to] X.

Definition 3.4. Let (X,[tau]) be a topological space and A be non-empty subset of X. let [G.sub.[alpha]] be arbitrary in [T.sub.[alpha]], then collection [[tau].sup.A.sub.[alpha]] = {[G.sub.[alpha]] [intersection] A : [G.sub.[alpha]] [member of] [[tau].sub.[alpha]]} is a topology on A, called the subspace or relative topology of topology [[tau].sub.[alpha]].

Theorem 3.5. If (X,[tau]) a disconnected space and (X,[[tau].sub.[alpha]]) is a topological space, then (X,[[tau].sub.[alpha]]) is [alpha]-[tau]-disconnected.

Proof. As (X,[tau]) disconnected and [[tau].sub.[alpha]] is finer than, hence by Theorem 3.1 (X,[tau]) is disconnected.

Theorem 3.6. Let (X,[tau]) and (X,[[tau].sub.[alpha]]) are spaces, then (X,[tau])is [alpha]-[tau]-disconnected if and only if there exists non-empty proper subset of X which is both [alpha]-open and [alpha]-closed.

Proof. Necessity: Let (X,[[tau].sub.[alpha]]) be [alpha]-[tau]-disconnected. Then, by Definition 3.3 there exist non-empty sets [G.sub.[alpha]] and [H.sub.[alpha]] in [[tau].sub.[alpha]] such that [G.sub.[alpha]] [intersection] [H.sub.[alpha]] is non-empty and [G.sub.[alpha]] [union] [H.sub.[alpha]] = X. Since [G.sub.[alpha]] [intersection] [H.sub.[alpha]] = [phi] and [H.sub.[alpha]] is open in [[tau].sub.[alpha]] show that [G.sub.[alpha]] = X - [H.sub.[alpha]], but it is [alpha]-closed. Hence [G.sub.[alpha]] is non-empty proper subset of X which is [alpha]-closed as well [alpha]-open.

Sufficiency: Suppose A is non-empty proper subset of X such that it is [alpha]-open as well [alpha]-closed. Now A is nonempty [alpha]-closed show that X - A is non-empty and [alpha]-open. Suppose B = X - A, then A [union] B = X and A [intersection] B = [phi]. Thus A and B are non-empty disjoint [alpha]-open as well as [alpha]-closed subset of X such that A [union] B = X. Consequently X is [alpha]-[tau] -disconnected.

Theorem 3.7. Every (X,[[tau].sub.[alpha]]) discrete space is [alpha]-[tau] -disconnected if the space contains more than one element.

Proof. Let (X,[[tau].sub.[alpha]]) be discrete space such that X = {a, b} contains more than one element. But [tau] is discrete topology so [tau] = {[phi],X,{a},{b}} and family of all [alpha]-open sets is [T.sub.[alpha]] = {[phi],X,{a},{b}}.

All [alpha]-closed sets are [phi],X,{a},{b}. Since {a} is non-empty proper subset of X which is both [alpha]-open and [alpha]-closed in X. Finally, we can say that (X ,[[tau].sub.[alpha]]) is [alpha]-t-disconnected by Theorem 3.6.

Theorem 3.8. A topological space (X,[[tau].sub.[alpha]]) is [alpha]-[tau]-connected if and only if one non-empty subset which is both [alpha]-open and [alpha]-closed in X is X itself.

Proof. Necessity: Assume that (X, [[tau].sub.[alpha]]) is [alpha]-[tau]-connected topological space. So, our assumption show that (X,[[tau].sub.[alpha]]) is not [alpha]-disconnected i.e. there does not exist a pair of non-empty disjoin [alpha]-open and [alpha]-closed A and B such that, A [union] B = X. This shows that there exist non-empty subsets (other than X) which are both and [alpha]-open and a closed in X.

Sufficiency: Suppose that (X,[[tau].sub.[alpha]]) is topological space such that the only non-empty subsets of X which is [alpha]-open as well as [alpha]-closed in X is X itself. By hypothesis, there does not exist a partition of the space X. Hence X is not [alpha]-[tau]-disconnected i.e. [alpha]-[tau] connected.

Theorem 3.9. Let A be a non-empty subset of topological space (X, r). Let TA be the relative topology on A , then A is [alpha]-[tau]-disconnected if and only if A is [alpha]-[tau]A -disconnected.

Proof. Necessity: Let A be a [alpha]-[tau]-disconnected and let [G.sub.[alpha]] [union] [H.sub.[alpha]] be a [alpha]-[tau]-disconnection on A. By Definition 3.3, there exists non-empty [G.sub.[alpha]] and [H.sub.[alpha]] in [[tau].sub.[alpha]] such that

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

Now by the definition of relative topology, if [G.sub.[alpha]] and [H.sub.[alpha]] in [[tau].sub.[alpha]], then there exist [G.sup.1.sub.[alpha]] and [H.sup.1.sub.[alpha]] in [[tau].sup.A.sub.[alpha]] such that [G.sup.1.sub.[alpha]] = A [intersection] [G.sub.[alpha]] and [H.sup.1.sub.[alpha]] = A [intersection] [H.sub.[alpha]].

Now by (1) [G.sup.1.sub.[alpha]] and [H.sup.1.sub.[alpha]] are non-empty. Hence A [intersection] [G.sup.1.sub.[alpha]] and A [intersection] [H.sup.1.sub.[alpha]] are non-empty.

Similarly by (2) and (3), we can say that (A [intersection] [G.sup.1.sub.[alpha]]) [intersection] (A [intersection] [H.sup.1.sub.[alpha]]) = [phi] and (A [intersection] [G.sup.1.sub.[alpha]]) [union] (A [intersection] [H.sup.1.sub.[alpha]]) = A respectively. Consequently A is [alpha]-[[tau].sup.A.sub.[alpha]] - disconnected.

Sufficiency: Suppose that A is [alpha]-[[tau].sup.A.sub.[alpha]]-disconnected and [M.sup.A.sub.[alpha]] [intersection] [N.sup.A.sub.[alpha]] is a [alpha]-[[tau].sup.A.sub.[alpha]]-disconnection on A. By definition, we can say that

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

Now (2) [??] there exists [M.sup.1.sub.[alpha]], [N.sup.1.sub.[alpha]] [member of] [[tau].sub.[alpha]] such that [M.sup.A.sub.[alpha]] = A [intersection] [M.sup.1.sub.[alpha]], [N.sup.1.sub.[alpha]] = A [intersection] [N.sup.1.sub.[alpha]]. But by (1) we can say that A [intersection] [M.sup.1.sub.[alpha]], A [intersection] [N.sup.1.sub.[alpha]] [not equal to] [phi]. Now (A [intersection] [M.sup.1.sub.[alpha]]) = A [intersection] (A [intersection] [M.sup.1.sub.[alpha]]) = (A [intersection] A) [intersection] [M.sup.1.sub.[alpha]] = A [intersection] [M.sup.1.sub.[alpha]].

Similarly we can say that A [intersection] [N.sup.A.sub.[alpha]] = A [intersection] [N.sup.1.sub.[alpha]].

Now (3) [??] (A [intersection] [M.sup.1.sub.[alpha]]) [intersection] (A [intersection] [N.sup.1.sub.[alpha]]) = [phi].

Similarly (4) [??] (A [intersection] [M.sup.1.sub.[alpha]]) [union] (A [intersection] [N.sup.1.sub.[alpha]]) = A.

Finally, we can say that A is [alpha]-[tau]-disconnected.

Theorem 3.10. The union of two non-empty [alpha]-[tau]-separate subsets of topological space (X,[[tau].sub.[alpha]]) is [alpha]-[tau]-disconnected.

Proof. Let A and B be [alpha]-[tau]-separate subsets of (X,[[tau].sub.[alpha]]). Then by definition 3.1, we can say that A and B non-empty.

A [intersection] [Cl.sub.[alpha]] (B), B [intersection] [Cl.sub.[alpha]] (A) = [phi];

A [intersection] B = [phi].

Let [alpha]-[Cl.sub.[alpha]] (A) = [G.sub.[alpha]] and [alpha]-[Cl.sub.[alpha]] (B) = [H.sub.[alpha]]. Then [Cl.sub.[alpha]] (A) and [Cl.sub.[alpha]] (B) are non-empty [alpha]-closed subsets of X which shows that [G.sub.[alpha]] and [H.sub.[alpha]] are non-empty [alpha]-open subsets of X.

Since,

[G.sub.[alpha]] [union] [H.sub.[alpha]] = (X - [Cl.sub.[alpha]] (A)) [union] (X - [Cl.sub.[alpha]] (B)) = X - [Cl.sub.[alpha]] (A) [intersection] [Cl.sub.[alpha]] (B)

we have

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

Similarly, (A [union] B) [intersection] [H.sub.[alpha]] = A. Now (1) shows that (A [union] B) [intersection] [G.sub.[alpha]], (A [union] B) [intersection] [H.sub.[alpha]] [not equal to] [phi].

Additionally, (3) shows that

[(A [union] B) [intersection] [H.sub.[alpha]]] [intersection] [(A [union] B) [intersection] [G.sub.[alpha]]] = 4 and

[(A [union] B) [intersection] [H.sub.[alpha]]] [union] [(A [union] B) [intersection] [G.sub.[alpha]]] = A [union] B

Finally, we can say that there exists [G.sub.[alpha]] and [H.sub.[alpha]] in [[tau].sub.[alpha]] such that

(A [union] B) [intersection] [G.sub.[alpha]], (A [unoin] B) [intersection] [H.sub.[alpha]] [not equal to] [phi]

[(A [union] B) [intersection] [H.sub.[alpha]]] [intersection] [(A [union] B) [intersection] [G.sub.[alpha]]] = 4 and

[(A [union] B) [intersection] [H.sub.[alpha]]] [union] [(A [union] B) [intersection] [G.sub.[alpha]]] = A [union] B.

So, [G.sub.[alpha]] [union] [H.sub.[alpha]] is a [alpha]-[tau]-disconnection of A [union] B. Hence A [union] B is [alpha]-[tau]-disconnected.

Theorem 3.11. Let (X,[tau]) and (X,[[tau].sub.[alpha]]) be topological spaces and A be a subset of X and let [G.sub.[alpha]] [union] [H.sub.[alpha]] be a [alpha]-[tau]-disconnection of A. Then A [intersection] [G.sub.[alpha]] and A [intersection] [H.sub.[alpha]] are [alpha]-[tau]-separate subsets of topological space (X,[[tau].sub.[alpha]]).

Proof. Let [G.sub.[alpha]] [union] [H.sub.[alpha]] be a given [alpha]-[tau] disconnection of subset A of (X,[[tau].sub.[alpha]]). To prove A [intersection] [G.sub.[alpha]] and A [intersection] [H.sub.[alpha]] are [alpha]-[tau]-separate subsets, we must show that

A [intersection] [G.sub.[alpha]] and A [intersection] H are non-empty;

[[Cl.sub.[alpha]](A [intersection] [G.sub.[alpha]])] [intersection] (A [intersection] [H.sub.[alpha]]) = [phi] and [[Cl.sub.[alpha]](A [intersection] [H.sub.[alpha]])] [intersection] (A [intersection] [G.sub.[alpha]]) = [phi];

Now by our assumption and definition, we can say that there exist [G.sub.[alpha]], [H.sub.[alpha]] [member of] [[tau].sub.[alpha]] such that

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

Evidently, (4) [??] (1).

To prove (2) suppose it is not possible i.e.

[Cl.sub.[alpha]](A [intersection] [G.sub.[alpha]]) [intersection] (A [intersection] [H.sub.[alpha]]) [not equal to] [phi].

Then, there exists x [member of] [Cl.sub.[alpha]](A [intersection] [G.sub.[alpha]]) [intersection] (A [intersection] [H.sub.[alpha]]) which implies that x [member of] [Cl.sub.[alpha]](A [intersection] [G.sub.[alpha]]) and x [member of] A, x [member of] [H.sub.[alpha]], that is (A [intersection] [G.sub.[alpha]]) [intersection] [H.sub.[alpha]] [not equal to] [phi]. Therefore (A [intersection] [G.sub.[alpha]]) [intersection] (A [intersection] [H.sub.[alpha]]) [not equal to] [phi]. But it is contrary to (5). Finally our assumption that is wrong.

Theorem 3.12. A subset Y of a topological space X is [alpha]-[tau]-disconnected if and only if it is union of two [alpha]-[tau]-separate sets.

Proof. Necessity: Suppose Y = A [union] B, where A and B are [alpha]-[tau]-separate sets of X.

By theorem 3.10, A [union] B is [alpha]-[tau]-disconnected. Hence, Y is [alpha]-[tau]-disconnected.

Sufficiency: Let Y is [alpha]-[tau]-disconnected. To prove that there exists two [alpha]-[tau]-separate subsets of A, B in X such that Y = A [union] B. By assumption, Y is [alpha]-[tau]-disconnected show that there exists a [alpha]-[tau]-disconnection [G.sub.[alpha]] [union] [H.sub.[alpha]] of Y. Therefore by Definition 3.3, we can say that there exists [G.sub.[alpha]] [H.sub.[alpha]] in [[tau].sub.[alpha]] such that

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

Since (Y [intersection] [G.sub.[alpha]]) and (Y[intersection] [H.sub.[alpha]]) are separated sets, if we write A = (Y [intersection] [G.sub.[alpha]]) and B = (Y [intersection] [H.sub.[alpha]]), then by (3) Y = A [union] B. Finally, we can say that there exist two [alpha]-[tau]-separate sets A and B in X such that Y = A [union] B.

Theorem 3.13. If Y is an [alpha]-[tau]-connected subset of topological space X such that Y [subset] A [union] B, where A and B is [alpha]-[tau]-connected, then Y [subset] A and Y [subset] B.

Proof. Since the inclusion Y [subset] A [union] B holds by the hypothesis we have (A [union] B) [intersection] Y = Y which yields that Y = (Y [intersection] A) [union] (Y [intersection] B). Now we want to prove that (Y [intersection] A), (Y [intersection] B) = 4. Suppose, (Y [intersection] A), (Y [intersection] B) [not equal to] [phi] Now, (Y [intersection] A) [intersection] [Cl.sub.[alpha]] (Y [intersection] B) [subset] (Y [intersection] A) [intersection] ([Cl.sub.[alpha]] (Y) [intersection] [Cl.sub.[alpha]] (B)) = (Y [intersection] [Cl.sub.[alpha]](Y)) [intersection] (A [intersection] [Cl.sub.[alpha]](B)) = (Y [intersection] (A [intersection] [Cl.sub.[alpha]](B)) = Y [intersection] 4 = 4 i e., (Y [intersection] A) [intersection] [Cl.sub.[alpha]](Y [intersection] B)) = [phi].

Similarly, we can prove that [Cl.sub.[alpha]] (Y [intersection] A) [intersection] (Y [intersection] B) = [phi]. Hence, from the above result we can say that Y is a union of two [alpha]-[tau]-separate sets (Y [intersection] A) and (Y [intersection] B). Consequently, Y is [alpha]-[tau]-disconnected. But this contradicts the fact that Y is [alpha]-[tau]-connected. Hence we can say that (Y [intersection] A), (Y [intersection] B) = 4. Now if (Y [intersection] A) = [phi], then Y = [phi] [union] (Y [intersection] B) = (Y [intersection] B) which gives that Y [subset] B. Similarly, we can prove that Y [subset] A if (Y [intersection] A) = [phi].

Conclusion

We have introduced new approach of separate sets, disconnected sets and connected sets called a-T~ separate sets, [alpha]-[tau]-disconnected sets and [alpha]-[tau]-connected sets of topological spaces with the help of [alpha]-open and [alpha]-closed sets and investigated their properties. The results of this paper will help to study various weak and strong form of connectedness and disconnectedness in topological spaces and it can be also applied for the study of topology in robotics, topological modeling and geographical information system.

Doi: 10.4025/actascitechnol.v37i3.17332

Acknowledgements

I am very thankful to referee and my colleague Dr. A. K. Srivastva for providing useful commnets for improvement of the paper.

Reference

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FARBER, M.; TABACHNIKOV, S.; YUZVINSKY, S. Topological robotics: Motion planning in projective spaces. International Mathematical Research Notices, v. 34, p. 1853-1870, 2003.

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MASHHOUR, A. S.; MONSEF, A. E.; DEEB, S. N. E. On Precontinuous and Weak Precontinuous Mapping. Proceedings of the Mathematical and Physical Society of Egypt, v. 53, p. 47-53, 1982.

MAHESHWARI, S. N.; JAIN, P. C. Some New Mappings. Mathematica, v. 24, n. 1-2, p. 53-55, 1982. NJASTAD, O. On Some [Cl.sub.[alpha]]sses of Nearly Open Sets. Pacific Journal of Mathematics, v. 15, n. 3, p. 961-970, 1965.

Received on May 23, 2012.

Accepted on September 28, 2012.

Sanjay Mishra

Department of Mathematics, School of Physical Sciences, Lovely Professional University, Punjab, India. E-mail: drsanjaymishra@rediffmail.com
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